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3 years ago

Prove that tan60+tan70-tan50+tan10=2sqrt3?

1 answer:

8 0

{tan(60) + tan(10)}/{1 - tan(60)*tan(10)} - {tan(60) - tan(10)}/{1 + tan(10)*tan(60)}

ii) Taking LCM & simplifying with applying tan(60) = √3, the above simplifies to: 

= 8*tan(10)/{1 - 3*tan²(10)} 

iii) So tan(70) - tan(50) + tan(10) = 8*tan(10)/{1 - 3*tan²(10)} + tan(10) 

= [8*tan(10) + tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)} 

= [9*tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)} 

= 3 [3*tan(10) - tan³(10)]/{1 - 3*tan²(10)} 

= 3*tan(30) = 3*(1/√3) = √3 [Proved] 

[Since tan(3A) = {3*tan(A) - tan³(A)}/{1 - 3*tan²(A)}, 
{3*tan(10) - tan³(10)}/{1 - 3*tan²(10)} = tan(3*10) = tan(30)]

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3 years ago

Question 8(Multiple Choice Worth 3 points)

Nastasia [14]

Answer: C) F(4) = 0

Detailed Explanation:

A zero/root of a polynomial is a value of the variable which makes the value of the polynomial to 0.

Therefore, when (x - 4) is a Factor :-
=> x - 4 = 0
=> x = 4

=> x = 4 is the Zero of the Polynomial.

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2 years ago

2/3x + 1/6 = -3/4x + 1

kupik [55]

$\frac{2}{3} x + \frac{1}{6} = -\frac{3}{4} x + 1$

First, simplify $\frac{2}{3} x$ to $\frac{2x}{3}$ / Your problem should look like: $\frac{2x}{3} + \frac{1}{6} = -\frac{3}{4} x + 1$
Second, simplify $\frac{3}{4} x$ to $\frac{3x}{4}$ / Your problem should look like: $\frac{2x}{3} + \frac{1}{6} = -\frac{3x}{4} + 1$
Third, multiply both sides by 12 (the LCM of 3 and 4) / Your problem should look like: $8x + 2 = -9x + 12$
Fourth, subtract 2 from both sides. / Your problem should look like: $8x = -9x + 12 - 2$
Fifth, simplify -9x + 12 - 2 to -9x + 10. / Your problem should look like: $8x = -9x + 10$
Sixth, add 9x to both sides. / Your problem should look like: $8x + 9x = 10$
Seventh, add 8x + 9x to get 17x. / Your problem should look like: $17x = 10$
Eighth, divide both sides by 17. / Your problem should look like: $x = \frac{10}{17}$

Answer as fraction: x = $\frac{10}{17}$
Answer as decimal: x = 0.5882

7 0

3 years ago

Tyler's mom purchased a savings bond for Tyler. The value of the savings bond increases by 4% each year. One year after it was p

Gemiola [76]

Answer:

The value of the bond when Tyler's mom purchased it was $150

Step-by-step explanation:

we know that

In this problem we have a exponential function of the form

$f(x)=a(b)^{x}$

where

a is the initial value (y-intercept)

b is the base

r is the rate

b=(1+r)

In this problem

r=4%=4/100=0.04

b=1+0.04=1.04

substitute

$f(x)=a(1.04)^{x}$

where

x is the number of years since the savings bond was purchased

f(x) is the value of the savings bond

For x=1

f(x)=$156

substitute

$156=a(1.04)^{1}$

Solve for a

$156=a(1.04)$

$a=\$150$

therefore

The value of the bond when Tyler's mom purchased it was $150

8 0

3 years ago

Complete the proof.

Harman [31]

2. Segment addition postulate
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3 years ago